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For convex polyhedra you have Steinitz's theorem characterizing them as the 3-connected planar graphs. My question is not about spheric tilings, but about periodic tilings of the euclidean plane. Is it here also the case that 3-connectivity corresponds with convexity?

It is easy to construct an example of a 2-connected periodic tiling that is not convex. My guess is that the symmetry group prohibits some tilings from being convex and thus there are also 3-connected periodic tilings that are not convex, but I can't seem to be able to construct a counter example to support this guess.

Is there a known counterexample for this? Or is it actually true? Any reference or hint would help.

edit: I was looking for a periodic tiling which has no equivariantly equivalent (i.e. topologically equivalent and the same symmetry group) tiling that is convex and that is 3-connected when you look at the segments as graph edges. My feeling is that in case of the euclidean plane there might be a tiling in which you can't convexify one face, without needing to make another one nonconvex to keep the same symmetry group.

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You are perhaps thinking of Tutte's "spring" theorem that every 3-connected graph has an embedding with convex faces. It indeed a corollary of the (earlier) Steinitz theorem, but Tutte's proof is of independent interest. Now, for periodic networks in the plane, you need to look at the "torus version" of the Tutte theorem. Start with this important paper which gives a version of the result and check some further refs therein.

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  • $\begingroup$ It may be that you answered a different question than what nvcleemp intended? Not certain. Rather than asking if a tiling could be convexified à la Tutte, I interpreted the question as asking whether there is a tiling with nonconvex pieces, which, if unaltered geometrically and with its segments interpreted as graph edges, is 3-connected. $\endgroup$
    – Joseph O'Rourke
    Commented Sep 17, 2011 at 0:20
  • $\begingroup$ Yes, maybe I forgot to add some things which were to obvious for me, because I've been working with those restrictions some time now. I was looking for a periodic tiling which has no equivariantly equivalent (i.e. topologically equivalent and the same symmetry group) tiling that is convex and that is 3-connected when you look at the segments as graph edges. I'll also add this part to the original question. $\endgroup$
    – nvcleemp
    Commented Sep 18, 2011 at 5:59
  • $\begingroup$ but thanks already for this answer: I'll certainly read the article. $\endgroup$
    – nvcleemp
    Commented Sep 18, 2011 at 6:03

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